# CBSE Class 11 Maths – Chapter 9 Sequences and Series- Study Materials

### Sequence

It means an arrangement of number in definite order according to some rule

Example
$2,4,6,8,10, 12,14………$
$1, 3, 5,9, 11, 13……..$
$1^2 +2^2 + 3^3 +……….$

Important point

• The various numbers occurring in a sequence are called its terms. The number of terms is called the length of the series
• Terms of a sequence are denoted generally by $a_1 , a_2, a_3, ….., a_n$ etc., The subscripts denote the position of the term. We can choose any other letter to denote it
• The nth term is the number at the nth position of the sequence and is denoted by an
• The nth term is also called the general term of the sequence
• A sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it of the type {1, 2, 3…k}.
• Many times is possible to express general term in terms of algebraic formula. But it may not be true in other cases. But we should be able to generate the terms of the sequence using some rules or theoretical scheme
Example
1. 0,2, 4,6,8…….
General Term an = 2n
2. 1, 1, 2, 3, 5, 8,..
This sequence does not have visible algebraic formula, but the sequence is generated by the recurrence relation given by
a1 =a2 =1
an =an-2 +an-1 , n > 2
This sequence is called Fibonacci sequence

Finite sequence
A sequence containing a finite number of terms is called Finite sequence
infinite sequence
A sequence is called infinite if it is not a finite sequence.

### Series

Let a1 , a2 , a3 ..be the sequence, then the sum expressed a1 + a2 +a3 + ……. is called series.

• A series is called finite series if it has got finite number of terms
• A series is called infinite series if it has got infinite terms
• Series are often represented in compact form, called sigma notation, using the Greek letter σ (sigma)
• so , a1 + a2 +a3 + …….an can be expressed as $\sum_{k=1}^{n}a_k$

Example
$1+2+3 + 4+ 5+….+100$
$= \sum_{k=1}^{n}k$

### Arithmetic Progression

An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant
Examples
1. $1,5,9,13,17….$
2. $1,2,3,4,5,…..$
1.The difference between any successive members is a constant and it is called the common difference of AP
2. If a1, a2,a3,a4,a5 are the terms in AP then
D=a2 -a1 =a3 – a2 =a4 – a3=a5 –a­4
3. We can represent the general form of AP in the form
a,a+d,a+2d,a+3d,a+4d..
Where a is first term and d is the common difference
4. Nth term of Arithmetic Progression is given by nthterm = a + (n – 1)d
5. Sum of nth item in Arithmetic Progression is given by
$S_n =\frac {n}{2}[a + (n-1)d]$
Or
$S_n =\frac {n}{2}[t_1+ t_n]$

Arithmetic Mean
The arithmetic mean A of any two numbers a and b is given by (a+b)/2 i.e. a, A, b are in AP
A -a = b- A or $A = \frac {(a+b)}{2}$
If we want to add n terms between A and B so that result sequence is in AP
Let A1, A2 , A3 , A4 , A5 …. An be the terms added between a and b
Then
b= a+ [(n+2) -1]d
or $d = \frac {(b-a)}{(n+1)}$
So terms will be $a+ \frac {(b-a)}{(n+1)}, a + 2 \frac {(b-a)}{(n+1)} ,……. a+ n \frac {(b-a)}{(n+1)}$

### Geometric Progressions

A sequence is said to be a geometric progression or G.P., if the ratio of any term to its preceding term is same throughout.
Examples
1. $2,4,8,16,32….$
2. $3,6,12,24,48,…..$

• The ratio of any term to its preceding term is same throughout. This constant factor is called the common ratio.
• If a1, a2,a3,a4,a5 are the terms in AP then
$r=\frac {a_2}{a_1} =\frac {a_3}{a_2} =\frac {a_4}{a_3}=\frac {a_5}{a_4}$
• We can represent the general form of G.P in the form
a,ar,ar2 ,ar3 ……
Where a is first term and r is the common ratio
• Nth term of Geometric Progression is given by nthterm = arn-1
• Sum of nth item inGeometric Progression is given by $S_n =a \frac {r^n -1}{r-1}$
Or
$S_n =a \frac {1-r^n }{ 1-r}$

Geometric Means
The Geometric mean G of any two numbers a and b is given by (ab)1/2 i.e. a, G, b are in G.P
$\frac {G}{a} = \frac {b}{G}$
or $G=\sqrt {ab}$
If we want to add n terms between A and B so that result sequence is in GP
Let G1, G2 , G3 , G4 , G5 …. Gn be the terms added between a and b
Then
$b= ar^{n+1}$
or $r= (\frac {b}{a})^{\frac {1}{n+1}}$
So terms will be $a(\frac {b}{a})^{\frac {1}{n+1}}, a(\frac {b}{a})^{\frac {2}{n+1}} ,……. a(\frac {b}{a})^{\frac {n}{n+1}}$
Relationship Between A.P and G.P
Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively.
Then
A ≥ G
And
$A – G = \frac {{\sqrt {a} – \sqrt {b}}^2}{2}$

### Sum of Special Series

Sum of first n natural numbers
$1 + 2 + 3 +….. + n = \frac {n(n+1}{2}$
$\sum_{k=1}^{n} n =\frac {n(n+1}{2}$
sum of squares of the first n natural numbers
$1^2+ 2^2+3^2 +…. + n^2 = \frac {n(n+1)(2n+1)}{6}$
$\sum_{k=1}^{n} n^2= \frac {n(n+1)(2n+1)}{6}$
Sum of cubes of the first n natural numbers
$1^3+ 2^3+3^3 +….. + n^3 = \frac {(n(n+1))^2}{4}$
$\sum_{k=1}^{n} n^3= \frac {(n(n+1))^2}{4}$

### Rules for finding the sum of Series

a. Write the nth term Tnof the series
b. Write the Tn in the polynomial form of n
$T_n= a n^3 + bn^2 + cn +d$
c. The sum of series can be written as
$\sum S = a \sum {n^3} + b \sum {n^2} + c \sum n + nd We already know the values if these standard from the formula given above and we can easily find the sum of the series Example Find the sum of the series$2^2 + 4^2 + 6^2 + …..(2n)^2}
Solution
Let nth term Tnof the series
Then
$T_n = (2n)^2 = 4n^2$
Now
$2^2 + 4^2 + 6^2 + …..(2n)^2 = \sum_{k=1}^{n} 4k^2 = 4 sum_{k=1}^{n} k^2 = 4 \frac {n(n+1)(2n+1)}{6}$
$=\frac {2n(n+1)(2n+1)}{3}$

### Method of Difference

Many times , nth term of the series can be determined. For example
5 + 11 + 19 + 29 + 41……
If the series is such that difference between successive terms are either in A.P or G.P, then we can the nth term using method of difference
$S_n = 5 + 11 + 19 + 29 + … + a_{n-1} + a_n$
or $S_n= 5 + 11 + 19 + … + + a_{n-1} + a_n$

On subtraction, we get
$0 = 5 + [6 + 8 + 10 + 12 + …(n – 1) terms] – a_n$
Here 6,8,10 is in A.P,So
$a_n = 5 + \frac {n-1}{2} [12 + (n-1)2]$
or $a_n= n^2 + 3n + 1$
Now it is easy to find the Sum of the series
$S_n = \sum_{k=1}^{n} {k^2 +3k +1}$
$=\frac {n(n+2)(n+4)}{3}$

### Sequences and Series Class 11 MCQs Questions with Answers

Question 1.
If a, b, c are in G.P., then the equations ax² + 2bx + c = 0 and dx² + 2ex + f = 0 have a common root if d/a, e/b, f/c are in
(a) AP
(b) GP
(c) HP
(d) none of these

Hint:
Given a, b, c are in GP
⇒ b² = ac
⇒ b² – ac = 0
So, ax² + 2bx + c = 0 have equal roots.
Now D = 4b² – 4ac
and the root is -2b/2a = -b/a
So -b/a is the common root.
Now,
dx² + 2ex + f = 0
⇒ d(-b/a)² + 2e×(-b/a) + f = 0
⇒ db2 /a² – 2be/a + f = 0
⇒ d×ac /a² – 2be/a + f = 0
⇒ dc/a – 2be/a + f = 0
⇒ d/a – 2be/ac + f/c = 0
⇒ d/a + f/c = 2be/ac
⇒ d/a + f/c = 2be/b²
⇒ d/a + f/c = 2e/b
⇒ d/a, e/b, f/c are in AP

Question 2.
If a, b, c are in AP then
(a) b = a + c
(b) 2b = a + c
(c) b² = a + c
(d) 2b² = a + c

Answer: (b) 2b = a + c
Hint:
Given, a, b, c are in AP
⇒ b – a = c – b
⇒ b + b = a + c
⇒ 2b = a + c

Question 3:
Three numbers form an increasing GP. If the middle term is doubled, then the new numbers are in Ap. The common ratio of GP is
(a) 2 + √3
(b) 2 – √3
(c) 2 ± √3
(d) None of these

Hint:
Let the three numbers be a/r, a, ar
Since the numbers form an increasing GP, So r > 1
Now, it is given that a/r, 2a, ar are in AP
⇒ 4a = a/r + ar
⇒ r² – 4r + 1 = 0
⇒ r = 2 ± √3
⇒ r = 2 + √3 {Since r > 1}

Question 4:
The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is
(a) n/(n+1)
(b) 1/(n+1)
(c) 1/n
(d) None of these

Hint:
Given series is:
S = (1/1·2) + (1/2·3) + (1/3·4) – ………………. 1/n.(n+1)
⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -……… (1/n – 1/(n+1))
⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)
⇒ S = 1 – 1/(n+1)
⇒ S = (n + 1 – 1)/(n+1)
⇒ S = n/(n+1)

Question 5:
If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then
(a) a, b, c are in AP
(b) a², b², c² are in AP
(c) 1/1, 1/b, 1/c are in AP
(d) None of these

Answer: (b) a², b², c² are in AP
Hint:
Given, 1/(b + c), 1/(c + a), 1/(a + b)
⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)
⇒ 2b² = a² + c²
⇒ a², b², c² are in AP

Question 6:
The sum of series 1/2! + 1/4! + 1/6! + ….. is
(a) e² – 1 / 2
(b) (e – 1)² /2 e
(c) e² – 1 / 2 e
(d) e² – 2 / e

Answer: (b) (e – 1)² /2 e
Hint:
We know that,
ex = 1 + x/1! + x² /2! + x³ /3! + x4 /4! + ………..
Now,
e1 = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ………..
e-1 = 1 – 1/1! + 1/2! – 1/3! + 1/4! + ………..
e1 + e-1 = 2(1 + 1/2! + 1/4! + ………..)
⇒ e + 1/e = 2(1 + 1/2! + 1/4! + ………..)
⇒ (e² + 1)/e = 2(1 + 1/2! + 1/4! + ………..)
⇒ (e² + 1)/2e = 1 + 1/2! + 1/4! + ………..
⇒ (e² + 1)/2e – 1 = 1/2! + 1/4! + ………..
⇒ (e² + 1 – 2e)/2e = 1/2! + 1/4! + ………..
⇒ (e – 1)² /2e = 1/2! + 1/4! + ………..

Question 7:
The third term of a geometric progression is 4. The product of the first five terms is
(a) 43
(b) 45
(c) 44
(d) none of these

Hint:
here it is given that T3 = 4.
⇒ ar² = 4
Now product of first five terms = a.ar.ar².ar³.ar4
= a5r10
= (ar2)5
= 45

Question 8:
Let Tr be the r th term of an A.P., for r = 1, 2, 3, … If for some positive integers m, n, we have Tm = 1/n and Tn = 1/m, then Tm n equals
(a) 1/m n
(b) 1/m + 1/n
(c) 1
(d) 0

Hint:
Let first term is a and the common difference is d of the AP
Now, Tm = 1/n
⇒ a + (m-1)d = 1/n ………… 1
and Tn = 1/m
⇒ a + (n-1)d = 1/m ………. 2
From equation 2 – 1, we get
(m-1)d – (n-1)d = 1/n – 1/m
⇒ (m-n)d = (m-n)/mn
⇒ d = 1/mn
From equation 1, we get
a + (m-1)/mn = 1/n
⇒ a = 1/n – (m-1)/mn
⇒ a = {m – (m-1)}/mn
⇒ a = {m – m + 1)}/mn
⇒ a = 1/mn
Now, Tmn = 1/mn + (mn-1)/mn
⇒ Tmn = 1/mn + 1 – 1/mn
⇒ Tmn = 1

Question 9.
The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is
(a) 2
(b) 4
(c) 6
(d) 8

Hint:
Let a and b are two numbers such that
a + b = 13/6
Let A1, A2, A3, ………A2n be 2n arithmetic means between a and b
Then, A1 + A2 + A3 + ………+ A2n = 2n{(n + 1)/2}
⇒ n(a + b) = 13n/6
Given that A1 + A2 + A3 + ………+ A2n = 2n + 1
⇒ 13n/6 = 2n + 1
⇒ n = 6

Question 10.
If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a/c, b/a, c/b are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) A.G.P.

Hint:
Given, equation is
ax² + bx + c = 0
Let p and q are the roots of this equation.
Now p+q = -b/a
and pq = c/a
Given that
p + q = 1/p² + 1/q²
⇒ p + q = (p² + q²)/(p² ×q²)
⇒ p + q = {(p + q)² – 2pq}/(pq)²
⇒ -b/a = {(-b/a)² – 2c/a}/(c/a)²
⇒ (-b/a)×(c/a)² = {b²/a² – 2c/a}
⇒ -bc²/a³ = {b² – 2ca}/a²
⇒ -bc²/a = b² – 2ca
Divide by bc on both side, we get
⇒ -c /a = b/c – 2a/b
⇒ 2a/b = b/c + c/a
⇒ b/c, a/b, c/a are in AP
⇒ c/a, a/b, b/c are in AP
⇒ 1/(c/a), 1/(a/b), 1/(b/c) are in HP
⇒ a/c, b/a, c/b are in HP

Question 11.
If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then
(a) a, b, c are in AP
(b) a², b², c² are in AP
(c) 1/1, 1/b, 1/c are in AP
(d) None of these

Answer: (b) a², b², c² are in AP
Hint:
Given, 1/(b + c), 1/(c + a), 1/(a + b)
⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)
⇒ 2b² = a² + c²
⇒ a², b², c² are in AP

Question 12.
The 35th partial sum of the arithmetic sequence with terms an = n/2 + 1
(a) 240
(b) 280
(c) 330
(d) 350

Hint:
The 35th partial sum of this sequence is the sum of the first thirty-five terms.
The first few terms of the sequence are:
a1 = 1/2 + 1 = 3/2
a2 = 2/2 + 1 = 2
a3 = 3/2 + 1 = 5/2
Here common difference d = 2 – 3/2 = 1/2
Now, a35 = a1 + (35 – 1)d = 3/2 + 34 ×(1/2) = 17/2
Now, the sum = (35/2) × (3/2 + 37/2)
= (35/2) × (40/2)
= (35/2) × 20
= 35 × 10
= 350

Question 13.
The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is
(a) 2
(b) 4
(c) 6
(d) 8

Hint:
Let a and b are two numbers such that
a + b = 13/6
Let A1, A2, A3, ………A2n be 2n arithmetic means between a and b
Then, A1 + A2 + A3 + ………+ A2n = 2n{(n + 1)/2}
⇒ n(a + b) = 13n/6
Given that A1 + A2 + A3 + ………+ A2n = 2n + 1
⇒ 13n/6 = 2n + 1
⇒ n = 6

Question 14.
The first term of a GP is 1. The sum of the third term and fifth term is 90. The common ratio of GP is
(a) 1
(b) 2
(c) 3
(d) 4

Hint:
Let first term of the GP is a and common ratio is r.
3rd term = ar²
5th term = ar4
Now
⇒ ar² + ar4 = 90
⇒ a(r² + r4) = 90
⇒ r² + r4 = 90
⇒ r² ×(r² + 1) = 90
⇒ r²(r² + 1) = 3² ×(3² + 1)
⇒ r = 3
So the common ratio is 3

Question 15.
The sum of AP 2, 5, 8, …..up to 50 terms is
(a) 3557
(b) 3775
(c) 3757
(d) 3575

Hint:
Given, AP is 2, 5, 8, …..up to 50
Now, first term a = 2
common difference d = 5 – 2 = 3
Number of terms = 50
Now, Sum = (n/2)×{2a + (n – 1)d}
= (50/2)×{2×2 + (50 – 1)3}
= 25×{4 + 49×3}
= 25×(4 + 147)
= 25 × 151
= 3775

Question 16.
If 2/3, k, 5/8 are in AP then the value of k is
(a) 31/24
(b) 31/48
(c) 24/31
(d) 48/31

Hint:
Given, 2/3, k, 5/8 are in AP
⇒ 2k = 2/3 + 5/8
⇒ 2k = 31/24
⇒ k = 31/48
So, the value of k is 31/48

Question 17.
The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is
(a) n/(n+1)
(b) 1/(n+1)
(c) 1/n
(d) None of these

Hint:
Given series is:
S = (1/1·2) + (1/2·3) + (1/3·4) – ……………….1/n.(n+1)
⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -………(1/n – 1/(n+1))
⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)
⇒ S = 1 – 1/(n+1)
⇒ S = (n + 1 – 1)/(n+1)
⇒ S = n/(n+1)

Question 18.
If the third term of an A.P. is 7 and its 7 th term is 2 more than three times of its third term, then the sum of its first 20 terms is
(a) 228
(b) 74
(c) 740
(d) 1090

Hint:
Let a is the first term and d is the common difference of AP
Given the third term of an A.P. is 7 and its 7th term is 2 more than three times of its third term
⇒ a + 2d = 7 ………….. 1
and
3(a + 2d) + 2 = a + 6d
⇒ 3×7 + 2 = a + 6d
⇒ 21 + 2 = a + 6d
⇒ a + 6d = 23 ………….. 2
From equation 1 – 2, we get
4d = 16
⇒ d = 16/4
⇒ d = 4
From equation 1, we get
a + 2×4 = 7
⇒ a + 8 = 7
⇒ a = -1
Now, the sum of its first 20 terms
= (20/2)×{2×(-1) + (20-1)×4}
= 10×{-2 + 19×4)}
= 10×{-2 + 76)}
= 10 × 74
= 740

Question 19.
If the sum of the first 2n terms of the A.P. 2, 5, 8, ….., is equal to the sum of the first n terms of the A.P. 57, 59, 61, ….., then n equals
(a) 10
(b) 12
(c) 11
(d) 13

Hint:
Given,
the sum of the first 2n terms of the A.P. 2, 5, 8, …..= the sum of the first n terms of the A.P. 57, 59, 61, ….
⇒ (2n/2)×{2×2 + (2n-1)3} = (n/2)×{2×57 + (n-1)2}
⇒ n×{4 + 6n – 3} = (n/2)×{114 + 2n – 2}
⇒ 6n + 1 = {2n + 112}/2
⇒ 6n + 1 = n + 56
⇒ 6n – n = 56 – 1
⇒ 5n = 55
⇒ n = 55/5
⇒ n = 11

Question 20.
If a is the A.M. of b and c and G1 and G2 are two GM between them then the sum of their cubes is
(a) abc
(b) 2abc
(c) 3abc
(d) 4abc

Hint:
Given, a is the A.M. of b and c
⇒ a = (b + c)
⇒ 2a = b + c ………… 1
Again, given G1 and G1 are two GM between b and c,
⇒ b, G1, G2, c are in the GP having common ration r, then
⇒ r = (c/b)1/(2+1) = (c/b)1/3
Now,
G1 = br = b×(c/b)1/3
and G1 = br = b×(c/b)2/3
Now,
(G1)³ + (G2)3 = b³ ×(c/b) + b³ ×(c/b)²
⇒ (G1)³ + (G2)³ = b³ ×(c/b)×( 1 + c/b)
⇒ (G1)³ + (G2)³ = b³ ×(c/b)×( b + c)/b
⇒ (G1)³ + (G2)³ = b² ×c×( b + c)/b
⇒ (G1)³ + (G2)³ = b² ×c×( b + c)/b ………….. 2
From equation 1
2a = b + c
⇒ 2a/b = (b + c)/b
Put value of(b + c)/b in eqaution 2, we get
(G1)³ + (G2)³ = b² × c × (2a/b)
⇒ (G1)³ + (G2)³ = b × c × 2a
⇒ (G1)³ + (G2)³ = 2abc

### Sequences and Series Class 11 MCQs Questions with Answers

Question 1.
If a, b, c are in G.P., then the equations ax² + 2bx + c = 0 and dx² + 2ex + f = 0 have a common root if d/a, e/b, f/c are in
(a) AP
(b) GP
(c) HP
(d) none of these

Hint:
Given a, b, c are in GP
⇒ b² = ac
⇒ b² – ac = 0
So, ax² + 2bx + c = 0 have equal roots.
Now D = 4b² – 4ac
and the root is -2b/2a = -b/a
So -b/a is the common root.
Now,
dx² + 2ex + f = 0
⇒ d(-b/a)² + 2e×(-b/a) + f = 0
⇒ db2 /a² – 2be/a + f = 0
⇒ d×ac /a² – 2be/a + f = 0
⇒ dc/a – 2be/a + f = 0
⇒ d/a – 2be/ac + f/c = 0
⇒ d/a + f/c = 2be/ac
⇒ d/a + f/c = 2be/b²
⇒ d/a + f/c = 2e/b
⇒ d/a, e/b, f/c are in AP

Question 2.
If a, b, c are in AP then
(a) b = a + c
(b) 2b = a + c
(c) b² = a + c
(d) 2b² = a + c

Answer: (b) 2b = a + c
Hint:
Given, a, b, c are in AP
⇒ b – a = c – b
⇒ b + b = a + c
⇒ 2b = a + c

Question 3:
Three numbers form an increasing GP. If the middle term is doubled, then the new numbers are in Ap. The common ratio of GP is
(a) 2 + √3
(b) 2 – √3
(c) 2 ± √3
(d) None of these

Hint:
Let the three numbers be a/r, a, ar
Since the numbers form an increasing GP, So r > 1
Now, it is given that a/r, 2a, ar are in AP
⇒ 4a = a/r + ar
⇒ r² – 4r + 1 = 0
⇒ r = 2 ± √3
⇒ r = 2 + √3 {Since r > 1}

Question 4:
The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is
(a) n/(n+1)
(b) 1/(n+1)
(c) 1/n
(d) None of these

Hint:
Given series is:
S = (1/1·2) + (1/2·3) + (1/3·4) – ………………. 1/n.(n+1)
⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -……… (1/n – 1/(n+1))
⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)
⇒ S = 1 – 1/(n+1)
⇒ S = (n + 1 – 1)/(n+1)
⇒ S = n/(n+1)

Question 5:
If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then
(a) a, b, c are in AP
(b) a², b², c² are in AP
(c) 1/1, 1/b, 1/c are in AP
(d) None of these

Answer: (b) a², b², c² are in AP
Hint:
Given, 1/(b + c), 1/(c + a), 1/(a + b)
⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)
⇒ 2b² = a² + c²
⇒ a², b², c² are in AP

Question 6:
The sum of series 1/2! + 1/4! + 1/6! + ….. is
(a) e² – 1 / 2
(b) (e – 1)² /2 e
(c) e² – 1 / 2 e
(d) e² – 2 / e

Answer: (b) (e – 1)² /2 e
Hint:
We know that,
ex = 1 + x/1! + x² /2! + x³ /3! + x4 /4! + ………..
Now,
e1 = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ………..
e-1 = 1 – 1/1! + 1/2! – 1/3! + 1/4! + ………..
e1 + e-1 = 2(1 + 1/2! + 1/4! + ………..)
⇒ e + 1/e = 2(1 + 1/2! + 1/4! + ………..)
⇒ (e² + 1)/e = 2(1 + 1/2! + 1/4! + ………..)
⇒ (e² + 1)/2e = 1 + 1/2! + 1/4! + ………..
⇒ (e² + 1)/2e – 1 = 1/2! + 1/4! + ………..
⇒ (e² + 1 – 2e)/2e = 1/2! + 1/4! + ………..
⇒ (e – 1)² /2e = 1/2! + 1/4! + ………..

Question 7:
The third term of a geometric progression is 4. The product of the first five terms is
(a) 43
(b) 45
(c) 44
(d) none of these

Hint:
here it is given that T3 = 4.
⇒ ar² = 4
Now product of first five terms = a.ar.ar².ar³.ar4
= a5r10
= (ar2)5
= 45

Question 8:
Let Tr be the r th term of an A.P., for r = 1, 2, 3, … If for some positive integers m, n, we have Tm = 1/n and Tn = 1/m, then Tm n equals
(a) 1/m n
(b) 1/m + 1/n
(c) 1
(d) 0

Hint:
Let first term is a and the common difference is d of the AP
Now, Tm = 1/n
⇒ a + (m-1)d = 1/n ………… 1
and Tn = 1/m
⇒ a + (n-1)d = 1/m ………. 2
From equation 2 – 1, we get
(m-1)d – (n-1)d = 1/n – 1/m
⇒ (m-n)d = (m-n)/mn
⇒ d = 1/mn
From equation 1, we get
a + (m-1)/mn = 1/n
⇒ a = 1/n – (m-1)/mn
⇒ a = {m – (m-1)}/mn
⇒ a = {m – m + 1)}/mn
⇒ a = 1/mn
Now, Tmn = 1/mn + (mn-1)/mn
⇒ Tmn = 1/mn + 1 – 1/mn
⇒ Tmn = 1

Question 9.
The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is
(a) 2
(b) 4
(c) 6
(d) 8

Hint:
Let a and b are two numbers such that
a + b = 13/6
Let A1, A2, A3, ………A2n be 2n arithmetic means between a and b
Then, A1 + A2 + A3 + ………+ A2n = 2n{(n + 1)/2}
⇒ n(a + b) = 13n/6
Given that A1 + A2 + A3 + ………+ A2n = 2n + 1
⇒ 13n/6 = 2n + 1
⇒ n = 6

Question 10.
If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a/c, b/a, c/b are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) A.G.P.

Hint:
Given, equation is
ax² + bx + c = 0
Let p and q are the roots of this equation.
Now p+q = -b/a
and pq = c/a
Given that
p + q = 1/p² + 1/q²
⇒ p + q = (p² + q²)/(p² ×q²)
⇒ p + q = {(p + q)² – 2pq}/(pq)²
⇒ -b/a = {(-b/a)² – 2c/a}/(c/a)²
⇒ (-b/a)×(c/a)² = {b²/a² – 2c/a}
⇒ -bc²/a³ = {b² – 2ca}/a²
⇒ -bc²/a = b² – 2ca
Divide by bc on both side, we get
⇒ -c /a = b/c – 2a/b
⇒ 2a/b = b/c + c/a
⇒ b/c, a/b, c/a are in AP
⇒ c/a, a/b, b/c are in AP
⇒ 1/(c/a), 1/(a/b), 1/(b/c) are in HP
⇒ a/c, b/a, c/b are in HP

Question 11.
If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then
(a) a, b, c are in AP
(b) a², b², c² are in AP
(c) 1/1, 1/b, 1/c are in AP
(d) None of these

Answer: (b) a², b², c² are in AP
Hint:
Given, 1/(b + c), 1/(c + a), 1/(a + b)
⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)
⇒ 2b² = a² + c²
⇒ a², b², c² are in AP

Question 12.
The 35th partial sum of the arithmetic sequence with terms an = n/2 + 1
(a) 240
(b) 280
(c) 330
(d) 350

Hint:
The 35th partial sum of this sequence is the sum of the first thirty-five terms.
The first few terms of the sequence are:
a1 = 1/2 + 1 = 3/2
a2 = 2/2 + 1 = 2
a3 = 3/2 + 1 = 5/2
Here common difference d = 2 – 3/2 = 1/2
Now, a35 = a1 + (35 – 1)d = 3/2 + 34 ×(1/2) = 17/2
Now, the sum = (35/2) × (3/2 + 37/2)
= (35/2) × (40/2)
= (35/2) × 20
= 35 × 10
= 350

Question 13.
The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is
(a) 2
(b) 4
(c) 6
(d) 8

Hint:
Let a and b are two numbers such that
a + b = 13/6
Let A1, A2, A3, ………A2n be 2n arithmetic means between a and b
Then, A1 + A2 + A3 + ………+ A2n = 2n{(n + 1)/2}
⇒ n(a + b) = 13n/6
Given that A1 + A2 + A3 + ………+ A2n = 2n + 1
⇒ 13n/6 = 2n + 1
⇒ n = 6

Question 14.
The first term of a GP is 1. The sum of the third term and fifth term is 90. The common ratio of GP is
(a) 1
(b) 2
(c) 3
(d) 4

Hint:
Let first term of the GP is a and common ratio is r.
3rd term = ar²
5th term = ar4
Now
⇒ ar² + ar4 = 90
⇒ a(r² + r4) = 90
⇒ r² + r4 = 90
⇒ r² ×(r² + 1) = 90
⇒ r²(r² + 1) = 3² ×(3² + 1)
⇒ r = 3
So the common ratio is 3

Question 15.
The sum of AP 2, 5, 8, …..up to 50 terms is
(a) 3557
(b) 3775
(c) 3757
(d) 3575

Hint:
Given, AP is 2, 5, 8, …..up to 50
Now, first term a = 2
common difference d = 5 – 2 = 3
Number of terms = 50
Now, Sum = (n/2)×{2a + (n – 1)d}
= (50/2)×{2×2 + (50 – 1)3}
= 25×{4 + 49×3}
= 25×(4 + 147)
= 25 × 151
= 3775

Question 16.
If 2/3, k, 5/8 are in AP then the value of k is
(a) 31/24
(b) 31/48
(c) 24/31
(d) 48/31

Hint:
Given, 2/3, k, 5/8 are in AP
⇒ 2k = 2/3 + 5/8
⇒ 2k = 31/24
⇒ k = 31/48
So, the value of k is 31/48

Question 17.
The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is
(a) n/(n+1)
(b) 1/(n+1)
(c) 1/n
(d) None of these

Hint:
Given series is:
S = (1/1·2) + (1/2·3) + (1/3·4) – ……………….1/n.(n+1)
⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -………(1/n – 1/(n+1))
⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)
⇒ S = 1 – 1/(n+1)
⇒ S = (n + 1 – 1)/(n+1)
⇒ S = n/(n+1)

Question 18.
If the third term of an A.P. is 7 and its 7 th term is 2 more than three times of its third term, then the sum of its first 20 terms is
(a) 228
(b) 74
(c) 740
(d) 1090

Hint:
Let a is the first term and d is the common difference of AP
Given the third term of an A.P. is 7 and its 7th term is 2 more than three times of its third term
⇒ a + 2d = 7 ………….. 1
and
3(a + 2d) + 2 = a + 6d
⇒ 3×7 + 2 = a + 6d
⇒ 21 + 2 = a + 6d
⇒ a + 6d = 23 ………….. 2
From equation 1 – 2, we get
4d = 16
⇒ d = 16/4
⇒ d = 4
From equation 1, we get
a + 2×4 = 7
⇒ a + 8 = 7
⇒ a = -1
Now, the sum of its first 20 terms
= (20/2)×{2×(-1) + (20-1)×4}
= 10×{-2 + 19×4)}
= 10×{-2 + 76)}
= 10 × 74
= 740

Question 19.
If the sum of the first 2n terms of the A.P. 2, 5, 8, ….., is equal to the sum of the first n terms of the A.P. 57, 59, 61, ….., then n equals
(a) 10
(b) 12
(c) 11
(d) 13

Hint:
Given,
the sum of the first 2n terms of the A.P. 2, 5, 8, …..= the sum of the first n terms of the A.P. 57, 59, 61, ….
⇒ (2n/2)×{2×2 + (2n-1)3} = (n/2)×{2×57 + (n-1)2}
⇒ n×{4 + 6n – 3} = (n/2)×{114 + 2n – 2}
⇒ 6n + 1 = {2n + 112}/2
⇒ 6n + 1 = n + 56
⇒ 6n – n = 56 – 1
⇒ 5n = 55
⇒ n = 55/5
⇒ n = 11

Question 20.
If a is the A.M. of b and c and G1 and G2 are two GM between them then the sum of their cubes is
(a) abc
(b) 2abc
(c) 3abc
(d) 4abc

Hint:
Given, a is the A.M. of b and c
⇒ a = (b + c)
⇒ 2a = b + c ………… 1
Again, given G1 and G1 are two GM between b and c,
⇒ b, G1, G2, c are in the GP having common ration r, then
⇒ r = (c/b)1/(2+1) = (c/b)1/3
Now,
G1 = br = b×(c/b)1/3
and G1 = br = b×(c/b)2/3
Now,
(G1)³ + (G2)3 = b³ ×(c/b) + b³ ×(c/b)²
⇒ (G1)³ + (G2)³ = b³ ×(c/b)×( 1 + c/b)
⇒ (G1)³ + (G2)³ = b³ ×(c/b)×( b + c)/b
⇒ (G1)³ + (G2)³ = b² ×c×( b + c)/b
⇒ (G1)³ + (G2)³ = b² ×c×( b + c)/b ………….. 2
From equation 1
2a = b + c
⇒ 2a/b = (b + c)/b
Put value of(b + c)/b in eqaution 2, we get
(G1)³ + (G2)³ = b² × c × (2a/b)
⇒ (G1)³ + (G2)³ = b × c × 2a
⇒ (G1)³ + (G2)³ = 2abc

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